1.) Given the following equations:

$3x-y=30\phantom{\rule{0ex}{0ex}}5x-3y=10$

What are the values of $x$ and $y$?

$3x-y=30\phantom{\rule{0ex}{0ex}}5x-3y=10$

What are the values of $x$ and $y$?

woowheedr
2022-07-06
Answered

1.) Given the following equations:

$3x-y=30\phantom{\rule{0ex}{0ex}}5x-3y=10$

What are the values of $x$ and $y$?

$3x-y=30\phantom{\rule{0ex}{0ex}}5x-3y=10$

What are the values of $x$ and $y$?

You can still ask an expert for help

asked 2022-07-03

How would you solve the following system of equations:

${x}^{2}+y=4\phantom{\rule{0ex}{0ex}}x+{y}^{2}=10$

I tried defining y in terms of x and then inserting in to the second equation:

$y=4-{x}^{2}\phantom{\rule{0ex}{0ex}}x+(4-{x}^{2}{)}^{2}=10$

Expand the second equation:

$x+16-8{x}^{2}+{x}^{4}=10$

Rearrange the terms:

${x}^{4}-8{x}^{2}+x+6=0$

${x}^{2}+y=4\phantom{\rule{0ex}{0ex}}x+{y}^{2}=10$

I tried defining y in terms of x and then inserting in to the second equation:

$y=4-{x}^{2}\phantom{\rule{0ex}{0ex}}x+(4-{x}^{2}{)}^{2}=10$

Expand the second equation:

$x+16-8{x}^{2}+{x}^{4}=10$

Rearrange the terms:

${x}^{4}-8{x}^{2}+x+6=0$

asked 2022-04-27

Solve $0.6-3.2x>-1.4-2.8x$ .

a.$x>5$

$b.x>10$ c.$5>x$

d.$x>1.62$

a.

d.

asked 2022-06-13

How to solve this simultaneous equations which involves exponents:

$\{\begin{array}{l}{2}^{x}-{2}^{y}=1\\ {4}^{x}-{4}^{y}=\frac{5}{3}\end{array}$

Simultaneously. From the second equation I factorized to get

$({2}^{x}+{2}^{y})({2}^{x}-{2}^{y})=\frac{5}{3}$

and from the first equation, since $({2}^{x}-{2}^{y})=1$, I deduced that

$({2}^{x}+{2}^{y})\cdot 1=\frac{5}{3}\to ({2}^{x}+{2}^{y})=\frac{5}{3}$

Now I am stuck on how to proceed.

$\{\begin{array}{l}{2}^{x}-{2}^{y}=1\\ {4}^{x}-{4}^{y}=\frac{5}{3}\end{array}$

Simultaneously. From the second equation I factorized to get

$({2}^{x}+{2}^{y})({2}^{x}-{2}^{y})=\frac{5}{3}$

and from the first equation, since $({2}^{x}-{2}^{y})=1$, I deduced that

$({2}^{x}+{2}^{y})\cdot 1=\frac{5}{3}\to ({2}^{x}+{2}^{y})=\frac{5}{3}$

Now I am stuck on how to proceed.

asked 2022-05-17

Given two integer variables $x$ and $y$. We are given that each integer variable $x$ and $y$ can't be greater than a given integer $z$.

The problem: We are given the proportions $a$ and $b$ such that $a+b=1$, $a=\frac{x}{z}$, and $b=\frac{y}{z}$. Is it possible to reverse solve $x$ and $y$?

The problem: We are given the proportions $a$ and $b$ such that $a+b=1$, $a=\frac{x}{z}$, and $b=\frac{y}{z}$. Is it possible to reverse solve $x$ and $y$?

asked 2022-09-07

Every $$3\times 3$$ skew symmetric matrix is singular. Because this is a skew symmetric matrix, $det(A)=det({A}^{T})=det(-A)=(-1{)}^{n}det(A)$, and when $$n$$ is odd $det(A)=-det(A)$, so $2det(A)=0$ and therefore $det(A)=0$. As such, the answer is "False" because it is only singular when n is odd.

asked 2022-07-15

How to solve this coupled 2nd order Differential equation of a double pendulum- Runge Kutta method

${\theta}_{1}^{\u2033}=\frac{-g(2{m}_{1}+{m}_{2})sin{\theta}_{1}-{m}_{2}gsin({\theta}_{1}-2{\theta}_{2})-2sin({\theta}_{1}-{\theta}_{2}){m}_{2}({\theta}_{2}^{\prime 2}{l}_{2}+{\theta}_{1}^{\prime 2}{l}_{1}cos({\theta}_{1}-{\theta}_{2})}{{l}_{1}(2{m}_{1}+{m}_{2}-{m}_{2}cos(2cos(2{\theta}_{1}-2{\theta}_{2})))}$

${\theta}_{2}^{\u2033}=\frac{2sin({\theta}_{1}-{\theta}_{2})({\theta}_{1}^{\prime}{l}_{1}({m}_{1}+{m}_{2})+g({m}_{1}+{m}_{2})cos{\theta}_{1}+{\theta}_{2}^{\prime 2}{l}_{2}{m}_{2}cos({\theta}_{1}-{\theta}_{2}))}{{l}_{2}(2{m}_{1}+{m}_{2}-{m}_{2}cos(2cos(2{\theta}_{1}-2{\theta}_{2})))}$

These are the equations, and How can I possibly solve this simultaneoouslt in RK4 method?

${m}_{1},{m}_{2}=$=masses of pendulum 1 and 2, ${\theta}_{1},{\theta}_{2}=$= angles formed by the pendulums, ${\theta}_{1}^{\prime}={\omega}_{1},{\theta}_{2}^{\prime}={\omega}_{2}$

${\theta}_{1}^{\u2033}=\frac{-g(2{m}_{1}+{m}_{2})sin{\theta}_{1}-{m}_{2}gsin({\theta}_{1}-2{\theta}_{2})-2sin({\theta}_{1}-{\theta}_{2}){m}_{2}({\theta}_{2}^{\prime 2}{l}_{2}+{\theta}_{1}^{\prime 2}{l}_{1}cos({\theta}_{1}-{\theta}_{2})}{{l}_{1}(2{m}_{1}+{m}_{2}-{m}_{2}cos(2cos(2{\theta}_{1}-2{\theta}_{2})))}$

${\theta}_{2}^{\u2033}=\frac{2sin({\theta}_{1}-{\theta}_{2})({\theta}_{1}^{\prime}{l}_{1}({m}_{1}+{m}_{2})+g({m}_{1}+{m}_{2})cos{\theta}_{1}+{\theta}_{2}^{\prime 2}{l}_{2}{m}_{2}cos({\theta}_{1}-{\theta}_{2}))}{{l}_{2}(2{m}_{1}+{m}_{2}-{m}_{2}cos(2cos(2{\theta}_{1}-2{\theta}_{2})))}$

These are the equations, and How can I possibly solve this simultaneoouslt in RK4 method?

${m}_{1},{m}_{2}=$=masses of pendulum 1 and 2, ${\theta}_{1},{\theta}_{2}=$= angles formed by the pendulums, ${\theta}_{1}^{\prime}={\omega}_{1},{\theta}_{2}^{\prime}={\omega}_{2}$

asked 2022-07-03

Solve the following system of differential equations

$\begin{array}{rl}\dot{x}& =2000-3xy-2x\\ \dot{y}& =3xy-6y\\ \dot{z}& =4y-2z\end{array}$

$\begin{array}{rl}\dot{x}& =2000-3xy-2x\\ \dot{y}& =3xy-6y\\ \dot{z}& =4y-2z\end{array}$